2 Answers
2

The answer is no. Consider the curve $C: y^2=x^2+x^3,$ and let $\pi:\widetilde C \to C$ be the blowup of $C$ at $0.$ Then the fibre $\pi^{-1}(0)=\{p,q\}$ consists of two points, while $\pi$ is an isomorphism everywhere else. Since $\widetilde C$ is a desinglularization of $C,$ it is isomorphic to the normalization of $C.$

The map $Y(k) \to X(k)$ can fail to be injective, or can fail to be surjective.

More precisely, if $x \in X(k)$ is a node, then the fibre of $Y$ over $x$ will
be a degree two reduced zero-dimensional variety, which will contain either two $k$-points
(in which case the map $Y(k) \to X(k)$ is not injective; this is an in Andrew's example), or no $k$-points, in which case $Y(k) \to X(k)$ will not be surjective.

The non-surjective case can't occur if $k$ is algebraically closed, since then the fibre over a node is just two copies of Spec $k$, but here is a case with $k = \mathbb R$ where it occurs:

Take $X$ to be the curve $y^2 = x^3 - x^2$. (To see that the normalization has no $\mathbb R$-valued points lying over the node, note that these points
correspond to the tangent directions to the two branches through the node. These tangent directions have the equation $x^2 + y^2 = 0$, whose linear factors are defined over $\mathbb C$, but not over $\mathbb R$.)

That clears things up pretty well. Fortunately, I don't really need the equality to hold for what I want to do. In fact, if $Y\to X$ is the normalization, it's clear that each smooth $k$-rational point on $X$ induces a (smooth) $k$-rational point on $Y$ by base change. Am I right?
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HarryJul 31 '12 at 8:42

@Harry: Dear Harry, Yes, that's right: the singular locus of $X$ is defined over $k$, so its complement is a smooth open curve defined over $k$, which is isomorphic to its preimage in $Y$; so the smooth $k$-points on $Y$ map isomorphically to the smooth $k$-points on $X$. Regards,
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Matt EJul 31 '12 at 12:13